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We are given two piecewise functions:

1. \( f(x) = \begin{cases} 
x + 5 & \text{if } x < -2 \\
x^2 + 2x + 3 & \text{if } x \geq -2 
\end{cases} \)

2. \( g(x) = \begin{cases} 
-6 & \text{if } x < -2 \\
(x + 1)^4 & \text{if } x \geq -2 
\end{cases} \)

We need to calculate \( f(3) + g(-2) \).

### Step 1: Calculate \( f(3) \)
Since \( 3 \geq -2 \), we use the second piece of \( f(x) \):
\[
f(3) = 3^2 + 2(3) + 3 = 9 + 6 + 3 = 18
\]

### Step 2: Calculate \( g(-2) \)
Since \( -2 \geq -2 \), we use the second piece of \( g(x) \):
\[
g(-2) = (-2 + 1)^4 = (-1)^4 = 1
\]

### Step 3: Add the results
\[
f(3) + g(-2) = 18 + 1 = 19
\]

Thus, \( f(3) + g(-2) = 19 \).

Would you like more details or have any questions about this solution?

Here are 5 related questions you can explore:
1. What happens if \( x = -2 \) in both \( f(x) \) and \( g(x) \)?
2. How would you find the derivative of the piecewise function \( f(x) \)?
3. What is the value of \( f(-3) + g(-3) \)?
4. Can you graph both \( f(x) \) and \( g(x) \) to explore their behavior visually?
5. How would you evaluate \( f'(x) \) and \( g'(x) \) at specific points?

**Tip:** Always check which condition applies in piecewise functions before solving!

This problem involves evaluating piecewise functions f(x) and g(x) at specific values. Learn how to solve f(3) + g(-2) using algebraic methods with step-by-step calculations. Suitable for students in Grades 9-11.

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